a 0    10  2

b       âm

a) Ta có : \(\left|3x+4\right|=2\left|2x-9\right|\)

=> \(\orbr{\begin{cases}3x+4=2\left(-2x+9\right)\\3x+4=2\left(2x-9\right)\end{cases}}\Rightarrow\orbr{\begin{cases}3x+4=-4x+18\\3x+4=4x-18\end{cases}}\Rightarrow\orbr{\begin{cases}7x=14\\-x=-22\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=22\end{cases}}\)

=> \(x\in\left\{2;22\right\}\)

b) Ta có : \(\left|10x+7\right|< 37\)

=> -37 < 10x + 7 < 37

=> -44 < 10x < 30

=> -4,4 < x < 3

Vậy -4,4 < x < 3

c) |3 - 8x| \(\le\)19

=> \(-19\le3-8x\le19\)

=> \(\hept{\begin{cases}3-8x\ge-19\\3-8x\le19\end{cases}}\Rightarrow\hept{\begin{cases}22\ge8x\\-16\le8x\end{cases}}\Rightarrow\hept{\begin{cases}x\le\frac{11}{4}\\x\ge-2\end{cases}}\Rightarrow-2\le x\le\frac{11}{4}\)

d) Ta có |x + 3| - 2x = |x - 4| (1)

Nếu x < -3

=> |x + 3| = -(x + 3) = -x - 3

=> |x - 4| = -(x - 4) = -x + 4

Khi đó (1) <=> -x - 3 - 2x = - x + 4

=> -3x - 3 = - x  + 4

=> -2x = 7

=> x = - 3,5 (tm)

Nếu \(-3\le x\le4\)

=> |x + 3| = x + 3

=> |x - 4| = -(x - 4) = -x + 4

Khi đó (1) <=> x + 3 - 2x = -x + 4

=> -x + 3 = -x + 4

=> 0x = 1 (loại)

Nếu x > 4

=> |x + 3| = x + 3

=> |x - 4| = x + 4

Khi đó (1) <=> x + 3 - 2x = x - 4

=> -x + 3 = x - 4

=> -2x = -7

=> x = 3,5 (loại)

Vậy x = -3,5

\(\left|3x+4\right|=2\left|2x-9\right|\)

\(\left|3x+4\right|\ge0\)

\(\left|2x-9\right|\ge0\Rightarrow2\left|2x-9\right|\ge0\)

\(\Rightarrow3x+4=2\left(2x-9\right)\)

\(3x+4=4x-18\)

\(3x=4x-14\)

\(x=14\)

\(\left|10x+7\right|\le37\)

\(\Rightarrow\left|10x+7\right|\le\left\{37;36;35;......;0\right\}\)

\(10x+7\le\left\{\pm37;\pm36;\pm35;.....0\right\}\)

Tự tính tiếp.C tương tự

\(\left|x+3\right|-2x=\left|x-4\right|\)

\(\left|x+3\right|=\left|x-4\right|+2x\)

\(\left|x+3\right|\ge0\)

\(\left|x-4\right|\ge0\)

\(\Rightarrow x+3=x-4+2x\)

\(x+3=3x-4\)

\(x=3x-7\)

\(x=\dfrac{7}{2}\)

Ta co : |3-8x| \(\le\)19

\(\Rightarrow\)3-8x=+19

TH1 : 3-8x=19

8x=-16

x=-2

TH2 : 3-8x=-19

8x=22

x=2,75

Nhớ k nha bạn

=>|8x-3|<=19

=>8x-3>=-19 và 8x-3<=19

=>8x>=-16 và 8x<=22

=>-2<=x<=11/4

a) \(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)\)

\(=\left[\left(2x^3+10x\right)+\left(x^4-25\right)\right]:\left(x^2+5\right)\)

\(=\left[2x\left(x^2+5\right)+\left(x^2-5\right)\left(x^2+5\right)\right]:\left(x^2+5\right)\)

\(=\left(x^2+5\right)\left(x^2+2x-5\right):\left(x^2+5\right)\)

\(=x^2+2x-5\)

1) \(A\left(x\right)=-5x^3+3x^4+\frac{5}{7}-8x^2-10x\)

\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

\(B\left(x\right)=-2x^4-\frac{2}{7}+7x^2+8x^3+6x\)

\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

2)       \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

      +

          \(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

\(A\left(x\right)+B\left(x\right)=x^4+3x^3-x^2-4x+\frac{3}{7}\)

                \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)

-

                \(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)

\(A\left(x\right)-B\left(x\right)=5x^4-13x^3-15x^2-16x+1\)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)